Here is a dissection based on Gerhard's [latest idea](http://mathoverflow.net/a/263267/6094). I didn't add this directly to his post because it is not identical to his $5$-piece dissection from unfoldings.
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[![TetraCubeGood][1]][1]
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The tetrahedron is unfolded to a $3.722 \times 1.611 = 6$ rectangle.
The cube is unfolded to two $3 \times 1$ rectangles, which are stacked to
$3 \times 2$. This leaves a $0.388 \times 3 = 1.164$ rectangle to cover a $0.722 \times 1.611 = 1.164$ rectangle. This is accomplished by the magic of
the Bolyai-Gerwien proof.

This appears to be $5$ pieces to form the $3.722 \times 1.611$ rectangle from the cube's two $3 \times 1$ rectangles. I consider this
a solution, starting from surface unfoldings, and have accepted Gerhard's answer.


  [1]: https://i.sstatic.net/B5YXC.jpg